The point of cookiemonster's formula is that you want to make
a equal to g: 9.81 m/s2. v is, of course, the linear speed. For a circle of radius R, so circumference 2&pi;R, moving with RPM &omega;, &omega revolutions per minute means covering 2&pi;&omega;R meters per minute or 2&pi;&omega;R/60= &pi;&omega;R/30 meters per second. That is: v= &pi;&omega;R/30 where v is in m/s and &omega is RPM. In terms of RPM, then
[tex] a= \frac{v^2}{R}= \frac{\pi^2\omega^2R^2}{900R}= \frac{\pi^2\omega^2R^2}{900}= 9.81\frac{m}{s^2} [/tex]

Strictly speaking, centrifugal force is a reaction force produced by the body on the rider. The only force acting on a body in uniform rotational motion is the centripetal force. However, the reaction force exerted by the ferris wheel on you makes you feel like you experience some kind of a centrifugal force.

It's from the viewpoint or perspective of the observer standing on the ground looking at the riders. The problem was, how fast the ferris wheel must rotate to make the riders feel weightless. The gravitational force on the riders is downwards, so at the top the counteracting force must be equal but opposite, i.e. upwards, away from the center of the ferris wheel. Do we call that centripetal? I am not saying that you are wrong, though. Just confusing. Of course from the viewpoint of the riders, the force they feel that is produced by gravity points upward. That's how the pressure on your seat of pants or on the soles of your feet feel. (I am not writing this for your benefit, Sickboy, but for others'). Hence, the force the riders feel that is produced by the ferris wheel is . . . (pause for an answer from the class - centripetal or centrifugal).